Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class A {

	void submit() {
		int r = nextInt();
		int c = nextInt();
		int[][] a = new int[r][c];
		for (int i = 0; i < r; i++) {
			for (int j = 0; j < c; j++) {
				a[i][j] = nextInt();
			}
		}
		
		int[] ansR = null;
		int[] ansC = null;
		int ansTot = Integer.MAX_VALUE;
		
		outer: for (int r0 = 0; r0 <= 500; r0++) {
			int[] nowR = new int[r];
			int[] nowC = new int[c];
			nowR[0] = r0;
			for (int j = 0; j < c; j++) {
				nowC[j] = a[0][j] - nowR[0];
			}
			for (int i = 1; i < r; i++) {
				nowR[i] = a[i][0] - nowC[0];
			}
			
			for (int i = 0; i < r; i++) {
				for (int j = 0; j < c; j++) {
					if (nowR[i] < 0 || nowC[j] < 0 || a[i][j] != nowR[i] + nowC[j]) {
						continue outer;
					}
				}
			}
			
			int nowTot = 0;
			for (int x : nowR) {
				nowTot += x;
			}
			
			for (int y : nowC) {
				nowTot += y;
			}
			
			if (nowTot < ansTot) {
				ansTot = nowTot;
				ansR = nowR;
				ansC = nowC;
			}
		}
		
		if (ansTot == Integer.MAX_VALUE) {
			out.println(-1);
		} else {
			out.println(ansTot);
			for (int i = 0; i < r; i++) {
				for (int j = 0; j < ansR[i]; j++) {
					out.println("row " + (i + 1));
				}
			}
			
			for (int i = 0; i < c; i++) {
				for (int j = 0; j < ansC[i]; j++) {
					out.println("col " + (i + 1));
				}
			}
		}
	}

	void preCalc() {

	}

	void stress() {

	}

	void test() {

	}

	A() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		preCalc();
		submit();
		//stress();
		//test();
		out.close();
	}

	static final Random rng = new Random();

	static int rand(int l, int r) {
		return l + rng.nextInt(r - l + 1);
	}

	public static void main(String[] args) throws IOException {
		new A();
	}

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			throw new RuntimeException(e);
		}
	}

	int nextInt() {
		return Integer.parseInt(nextToken());
	}

	long nextLong() {
		return Long.parseLong(nextToken());
	}

	double nextDouble() {
		return Double.parseDouble(nextToken());
	}
}